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4 years ago

13

Fusion and fission are both processes that involve which fundamental force? to ask? A. gravitational B. weak nuclear C. electrom

agnetic D. strong nuclear

1 answer:

marin [14]4 years ago

5 0

Fusion & Fission are both processes that involve D. Strong Nuclear Fundamental forces
"D". is the answer

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Humans are capable of vastly more complex thoughts and feelings due to the ever-adapting web of connection among?

salantis [7]

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3 years ago

A 1.2 kg block is held at rest against the spring with a force constant k= 730 N/m. Initially, the spring is compressed a distan

Nesterboy [21]

Answer:

Compression distance: $d \approx 0.102\,m$

Explanation:

According to this statement, we know that system is non-conservative due to the rough patch. By Principle of Energy Conservation and Work-Energy Theorem, we have the following expression that represents the system having a translational kinetic energy ( $K$ ), in joules, at the expense of elastic potential energy ( $U$ ), in joules, and overcoming work losses due to friction ( $W_{l}$ ), in joules:

$K + W_{l} = U$ (1)

By definitions of translational kinetic and elastic potential energies and work losses due to friction, we expand the equation described above:

$\frac{1}{2}\cdot m \cdot v^{2} +\mu\cdot m\cdot g \cdot s = \frac{1}{2} \cdot k \cdot d^{2}$ (2)

Where:

$m$ - Mass of the block, in kilograms.

$v$ - Final velocity of the block, in meters per second.

$\mu$ - KInetic coefficient of friction, no unit.

$g$ - Gravitational acceleration, in meters per square second.

$s$ - Width of the rough patch, in meters.

$k$ - Spring constant, in newtons per meter.

$d$ - Compression distance, in meters.

If we know that $m = 1.2\,kg$ , $v = 2.3\,\frac{m}{s}$ , $\mu = 0.44$ , $g = 9.807\,\frac{m}{s^{2}}$ , $s = 0.05\,m$ and $k = 730\,\frac{N}{m}$ , then the compression distance of the spring is:

$\frac{1}{2}\cdot m \cdot v^{2} +\mu\cdot m\cdot g \cdot s = \frac{1}{2} \cdot k \cdot d^{2}$

$m\cdot v^{2} + 2\cdot m\cdot g \cdot s = k\cdot d^{2}$

$d = \sqrt{\frac{m\cdot (v^{2}+2\cdot g\cdot s)}{k} }$

$d \approx 0.102\,m$

4 0

3 years ago

With what velocity Must a 0.53 kg softball be moving to be equal to the momentum of a 0.31 kg baseball moving at 21 m/s

vladimir2022 [97]

Answer:12.28m/s

Explanation:

momentum of baseball =mass of baseball x velocity of baseball

Momentum of baseball =0.31x21

Momentum of baseball =6.51kgm/s

For a softball to have same momentum with the baseball we can say :momentum of baseball =mass of softball x velocity of softball

6.51=0.53 x velocity of softball

Velocity of softball =6.51/0.53

Velocity of softball =12.28m/s

3 0

4 years ago

On Ramesh’s13th birthday, his father invited all his friends and their relatives. It was a big party with lots of food and DJs.

BaLLatris [955]

I go inside and take a nap normally that will help maybe play music

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3 years ago

Pleaseeeeeee helppppp this is mah last question... thank u (. ^ ᴗ ^. )

stepan [7]

Answer:

44.1 m

Explanation:

Consider the displacement being x:

x = (g•t²)/2

x = (9.8•3²)/2

x = (9.8•9)/2

x = 88.2/2

x = 44.1 m

4 0

3 years ago

Read 2 more answers